a) Let n be the pumping lemma constant. Then if L is regular, PL implies that s can be decomposed into xyz, |y| > 0, |xy| ≤n, such that xy i z is in L for all i ≥0.

Since the length of xy ≤n, y consists of all b's Then xy 2 z = anbncn, where the length of of y = j. We know j > 0 so the length of the pumped string contains at as many a's as b's as c's, and is not in L. This is a Contradiction L = {w :| n a (w) = n b (w) = nc(w)}

b)

• Let n be the pumping lemma constant. Then if L is regular, PL implies that s = anbncm can be decomposed into xyz, |y| > 0, |xy| ≤n, such thatxy i z is in L for all i ≥0.

• Since the length of xy ≤n, there are three ways to partition s:

1. y consists of all a's

Pumping y will lead to a string with more than n a's -- not in L

2. y consists of all b's

Pumping y will lead to a string with more than m b's, and leave

the number of c's untouched, such that there are no longer 2n more c's than b's -- not in L

3. y consists of a's and b's

Pumping y will lead to a string with b's before a's, -- not in L

• There is no way to partition anbncm that pumped strings are still in L.